– Homework 5 – tran – (52970)
This print-out should have 20 questions.
Multiple-choice questions may continue on
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before answering.
001
1. y =
′
2. y =
3. y ′ =
′
4. y =
5. y ′ =
6. y ′ =
2. f ′ (x) = 18 sin(18x + 5)
3. f ′ (x) = −18 sin(18x + 5) correct
10.0 points
Find the derivative of y when
√
√
√
y = 8 sin( x) − 10 x cos( x) .
′
1
sin(√x) √
5 cos( x) − 9
x
√
sin( x) √
√
4 sin( x) + 9
x
√
√
sin( x)
4 cos( x) + √
x
cos(√x) √
√
5 sin( x) − 9
x
√
√
cos( x)
5 sin( x) − √
correct
x
√
√
cos( x)
4 sin( x) + √
x
√
Explanation:
By the Product and Chain Rules,
cos(√x) cos(√x) ′
√
√
y =4
−5
x
x
√x sin(√x) √
+5
.
x
Consequently,
√
√
cos( x)
′
y = 5 sin( x) − √
.
x
002 10.0 points
Find the derivative of f when
f (x) = cos(9x + 7) cos(9x − 2)
− sin(9x + 7) sin(9x − 2) .
1. f ′ (x) = sin(9x + 7) cos(9x − 2)
4. f ′ (x) = cos(9x + 7) sin(9x − 2)
5. f ′ (x) = −9 sin(18x + 5)
Explanation:
Since
f (x) = cos(9x + 7) cos(9x − 2)
− sin(9x + 7) sin(9x − 2)
h
i
= cos (9x + 7) + (9x − 2)
= cos(18x + 5) ,
we see that
f ′ (x) = −18 sin(18x + 5) .
003
10.0 points
Find f ′ (x) when
f (x) = 2 sec2 x − tan2 x .
1. f ′ (x) = 2 tan2 sec x
2. f ′ (x) = 6 sec2 x tan x
3. f ′ (x) = −2 sec2 x tan x
4. f ′ (x) = −2 tan2 sec x
5. f ′ (x) = 2 sec2 x tan x correct
6. f ′ (x) = 6 tan2 sec x
Explanation:
Since
d
sec x = sec x tan x,
dx
d
tan x = sec2 x,
dx
– Homework 5 – tran – (52970)
Thus
the Chain Rule ensures that
f ′ (x) = 4 sec2 x tan x − 2 tan x sec2 x .
Consequently,
f ′ (x) = 2 sec2 x tan x .
004
10.0 points
Determine f ′ (x) when
r
x−1
f (x) =
.
x+1
2
1. f (x) =
1/2
(x − 1) (x + 1)3/2
′
2. f ′ (x) = −
3. f ′ (x) =
5. f ′ (x) =
6. f ′ (x) =
1
f (x) =
2
′
x+1
x−1
1/2
·
2
.
(x + 1)2
Consequently,
f ′ (x) =
1
(x − 1)1/2 (x +
005
1)3/2
Find the derivative of f when
f (x) = 2 tan 4x cos3 4x .
2
3. f ′ (x) = 8 cos 4x (1 − 3 cos2 4x)
(x
− 1)3/2 (x +
1)1/2
1
correct
1/2
(x − 1) (x + 1)3/2
Explanation:
To apply the Chain Rule it’s simpler to
write
r
1/2
x−1
x−1
f (x) =
=
.
x+1
x+1
For then,
2. f ′ (x) = 8 cos 4x (1 + 3 sin2 4x)
4. f ′ (x) = 2 cos 4x(1 − 3 cos2 4x)
5. f ′ (x) = 8 cos 4x (1 − 3 sin2 4x) correct
Explanation:
Using the fact that
d
1
tan x =
,
dx
cos2 x
x−1
x+1
1
=
2
x+1
x−1
−1/2
1/2
d x−1
dx x + 1
d x−1
.
dx x + 1
But by the Quotient Rule,
(x + 1) − (x − 1)
d x−1
=
dx x + 1
(x + 1)2
2
=
.
(x + 1)2
d
cos x = − sin x,
dx
together with the Chain rule, we obtain
f ′ (x) =
1
f (x) =
2
.
10.0 points
1. f ′ (x) = 2 cos 4x(3 sin2 4x − 1)
2
3/2
(x − 1) (x + 1)1/2
′
1
1/2
(x − 1) (x + 1)3/2
1
3/2
(x − 1) (x + 1)1/2
4. f ′ (x) = −
2
8
cos3 4x
2
cos 4x
− 24 tan 4x cos2 4x sin 4x.
Consequently,
f ′ (x) = 8 cos 4x (1 − 3 sin2 4x) .
Notice that the problem slightly simpler if
we observe that
tan 4x =
sin 4x
,
cos 4x
– Homework 5 – tran – (52970)
3
Determine f ′ (x) when
so that
√
f (x) = 2 sin 4x cos2 4x,
and then differentiate this function using the
known derivatives of sin 4x and cos 4x.
006
10.0 points
′
Find the value of F (2) when
F (x) = f (g(x))
f (x) = e
2x+3
.
√
2e 2x+3
1. f ′ (x) = √
2x + 3
√
2. f ′ (x) = 2e
√
3. f ′ (x) = e
2x+3
2x+3
√
2x + 3
√
and
′
g(2) = 5,
g (2) = 3 ,
f ′ (2) = 2,
f ′ (5) = 4 .
′
1. F (2) = 8
1 e 2x+3
4. f (x) = √
2 2x + 3
′
√
e
5. f ′ (x) = √
2x+3
2x + 3
correct
Explanation:
By the chain rule
′
2. F (2) = 11
√
′
f (x) = e
3. F ′ (2) = 12 correct
2x+3
√
4. F ′ (2) = 10
e
= √
d√
2x + 3
dx
2x+3
2x + 3
.
5. F ′ (2) = 9
008
Explanation:
By the Chain Rule,
If y is defined implicitly by
F ′ (x) = f ′ (g(x))g ′(x) .
Thus
′
′
′
F (2) = f (g(2))g (2)
= f ′ (5)g ′ (2) .
Consequently, when
g(2) = 5,
g ′ (2) = 3 ,
f ′ (2) = 2,
f ′ (5) = 4 ,
we see that
F ′ (2) = 12 .
Notice that the value of f ′ (2) was not needed.
007
10.0 points
10.0 points
4y 2 − xy − 6 = 0 ,
find the value of dy/dx at (10, 3).
dy 3
1.
=
correct
dx (10, 3)
14
dy 3
2.
= −
dx (10, 3)
14
dy 1
3.
=
dx (10, 3)
5
dy 2
4.
= −
dx (10, 3)
7
2
dy =
5.
dx (10, 3)
7
Explanation:
– Homework 5 – tran – (52970)
Differentiating implicitly with respect to x
we see that
8y
dy
dy
−y−x
= 0.
dx
dx
Thus
dy
y
=
.
dx
8y − x
keywords: implicit differentiation, Folium of
Descartes, derivative,
010
Find
dy
2 − sec2 (x + y)
=
dx
sec2 (x + y) + 3
2.
dy
3 + sec2 (x + y)
=
dx
sec2 (x + y) + 2
3.
dy
2 − sec2 (x + y)
=
dx
sec2 (x + y) − 3
4.
3 − sec2 (x + y)
dy
=
dx
sec2 (x + y) − 2
dy
3y + x2
=
dx
2y 2 + 3x
5.
3 − sec2 (x + y)
dy
=
correct
dx
sec2 (x + y) + 2
3y − x2
dy
=
dx
2y 2 − 3x
6.
dy
2 + sec2 (x + y)
=
dx
sec2 (x + y) − 3
Find
10.0 points
dy
when
dx
x3 − 2y 3 − 9xy − 1 = 0 .
2.
3.
dy
x2 + 3y
=
dx
2y 2 − 3x
dy
x2 − 3y
4.
=
correct
dx
2y 2 + 3x
5.
tan(x + y) = 3x − 2y .
1.
009
1.
10.0 points
dy
when
dx
At (10, 3), therefore,
dy 3
.
=
dx (10, 3)
14
4
x2 − 3y
dy
=
dx
2y 2 − 3x
Explanation:
We use implicit differentiation. For then
3x2 − 6y 2
dy
dy
− 9y − 9x
= 0,
dx
dx
which after solving for dy/dx and taking out
the common factor 3 gives
dy
3 (x2 − 3y) − 3 (2y 2 + 3x) = 0 .
dx
Explanation:
Differentiating implicitly with respect to x,
we see that
dy dy
2
= 3−2 .
sec (x + y) 1 +
dx
dx
After rearranging, this becomes
dy 2
sec (x + y) + 2 = 3 − sec2 (x + y) .
dx
Consequently,
dy
3 − sec2 (x + y)
=
.
dx
sec2 (x + y) + 2
Consequently,
dy
x2 − 3y
=
.
dx
2y 2 + 3x
keywords:
011
10.0 points
– Homework 5 – tran – (52970)
Now we can use the point-slope formula to
find an equation for the tangent line:
y−2 =
9
(x − 1) .
4
2. f ′ (x) = √
9
1
y = x−
.
4
4
013
3. f ′ (x) = √
10.0 points
4. f ′ (x) = √
Find dy/dx when
e2y = 10x2 + 5y 2 .
1.
dy
2x
=
2
dx
2x + 5y 2 − 5y
2.
5x
dy
=
2
dx
2x + y 2 + y
3.
2x
dy
=
2
dx
2x + y 2 + y
5. f ′ (x) = √
6. f ′ (x) = √
3
1 − x2
6
4 − x2
2
4 − x2
6
1 − x2
together with the Chain Rule shows that
3
1
′
f (x) = p
.
2
2
1 − (x/2)
Consequently,
Explanation:
Differentiating
f ′ (x) = √
e2y = 10x2 + 5y 2
015
implicitly with respect to x we see that
dy
dy
= 20x + 10y ,
dx
dx
so
dy
10x
= 2y
.
dx
e − 5y
dy
2x
=
2
dx
2x + y 2 − y
10.0 points
.
3
.
4 − x2
10.0 points
Find the derivative of
f (x) = tan−1 (e3x ) .
1. f ′ (x) =
1
1 + e6x
2. f ′ (x) =
3
1 + e6x
Thus
014
3
correct
4 − x2
d
1
arcsin(x) = √
,
dx
1 − x2
2x
dy
=
correct
2
dx
2x + y 2 − y
2e2y
2
1 − x2
Explanation:
Use of
5x
dy
=
4.
2
dx
2x + y 2 − y
5.
Determine the derivative of
x
.
f (x) = 3 arcsin
2
1. f ′ (x) = √
After simplification this becomes
6
3. f ′ (x) = √
4. f ′ (x) =
3e3x
1 − e6x
e3x
1 + e6x
– Homework 5 – tran – (52970)
Consequently,
3
5. f (x) = √
1 − e6x
′
6. f ′ (x) = √
7. f ′ (x) = √
8. f ′ (x) =
f ′ (θ) = 4 cot 4θ .
e3x
1 − e6x
017
1
1 − e6x
3e3x
correct
1 + e6x
d ax
e = aeax ,
dx
the Chain Rule ensures that
3e3x
f (x) =
.
1 + e6x
′
016
10.0 points
Find the derivative of f when
f (θ) = ln (sin 4θ) .
1. f ′ (θ) = cot 4θ
2. f ′ (θ) =
Find the derivative of f when
p
f (x) = 2 ln(x − x2 − 9), (x > 3) .
2. f ′ (x) = √
4. f ′ (x) = − √
6. f ′ (x) = √
5. f ′ (θ) = − tan 4θ
1
cos 4θ
Explanation:
By the Chain Rule,
f ′ (θ) =
4 cos 4θ
1
d
(sin 4θ) =
.
sin(4θ) dθ
sin 4θ
−9
4
x2 − 9
1
x2 − 9
1
x2 − 9
2
x2 − 9
2
x
√
1− √
x − x2 − 9
x2 − 9
f ′ (x) =
= −√
018
4. f (θ) = 4 tan 4θ
correct
Explanation:
By the Chain Rule
4
sin 4θ
′
−9
4
3. f ′ (x) = − √
5. f ′ (x) = √
2
x2
x2
3. f ′ (θ) = 4 cot 4θ correct
6. f ′ (θ) =
10.0 points
1. f ′ (x) = − √
Explanation:
Since
1
d
tan−1 x =
,
dx
1 + x2
7
2
x2 − 9
10.0 points
Find the derivative of
s
f (x) = ln
1. f ′ (x) =
2. f ′ (x) =
.
1 + 2x2
.
1 − 2x2
4x
correct
1 − 4x4
2x
1 − 4x2
3. f ′ (x) = −
4x
1 − 4x4
– Homework 5 – tran – (52970)
4. f ′ (x) = −
4x3
1 − 4x4
6. y ′ = xy(2 ln x − 1)
7. y ′ =
2x3
5. f (x) = −
1 − 4x2
′
Explanation:
Properties of logs ensure that
s
1 + 2x2
ln
1 − 2x2
q
q
2
= ln (1 + 2x ) − ln (1 − 2x2 )
1
ln(1 + 2x2 ) − ln(1 − 2x2 ) .
2
By the Chain rule, therefore,
4x 1 4x
f ′ (x) =
+
2 1 + 2x2 1 − 2x2
= 2x
(1 − 2x2 ) + (1 + 2x2 ) 1 − 4x2
Consequently,
f ′ (x) =
019
4x
1 − 4x4
.
10.0 points
Determine y ′ when
y = x1/x .
1. y ′ = −xy(2 ln x + 1)
Explanation:
After taking natural logs we see that
ln y =
3. y ′ = y(ln x + 1)
4. y ′ = −y(ln x + 1)
y
(ln x − 1) correct
x2
1
ln x .
x
Thus by implicit differentiation,
1 dy
1
1
= − 2 ln x + 2 .
y dx
x
x
Consequently,
y′ = −
020
Find
.
y
(ln x − 1) .
x2
10.0 points
dy
when
dx
y ln x + y 2 = x .
1.
dy
x+y
=
dx
x ln x − 2xy
2.
1−y
dy
=
dx
ln x + 2y
3.
dy
1+y
=
dx
ln x + 2y
4.
dy
x−y
=
dx
ln x − 2xy
5.
dy
1 + xy
=
dx
x ln x − 2y
6.
dy
x−y
=
correct
dx
x ln x + 2xy
2. y ′ = xy(2 ln x + 1)
5. y ′ = −
y
(ln x − 1)
x2
8. y ′ = −y(2 ln x − 1)
4x3
6. f ′ (x) =
1 − 4x4
=
8
Explanation:
Differentiating implicitly with respect to x
we see that
1
dy
dy
y
+ (ln x)
+ 2y
= 1,
x
dx
dx
– Homework 5 – tran – (52970)
which after rearrangement becomes
(ln x + 2y)
y
dy
= 1− .
dx
x
Consequently,
dy
x−y
=
dx
x ln x + 2xy
.
9